Question

Ignoring early effect, if C1 is the total capacitance tied to the emitter, what is the input pole of a simple C.B. stage?

(a) 1/gm * C1

(b) 2/gm * C1

(c) gm * C1

(d) gm * 2C1

This question was posed to me during an interview for a job.

My query is from Effect of Various Capacitors on Frequency Response topic in section Transistor Frequency Response of Analog Circuits

Answer 1

Right choice is (a) 1/gm * C1

Best explanation: The resistance looking into the emitter of the B.J.T. is 1/gm. The capacitance connected to the input node is C1 (as mentioned). The inverse product of these two provides us the input pole of the C.B. stage.